The coefficient of x^5 in the expansion of le | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The general term of $\left(2 x^3-\frac{1}{3 x^2}\right)^5$ is given by $T_{r+1} = { }^5 C_r (2 x^3)^{5-r} (-\frac{1}{3 x^2})^r$. Simplifying the e

Vedclass · Accepted Answer

$\frac{80}{9}$

Vedclass · Answer

(C) The general term of $\left(2 x^3-\frac{1}{3 x^2}ight)^5$ is given by $T_{r+1} = { }^5 C_r (2 x^3)^{5-r} (-\frac{1}{3 x^2})^r$. Simplifying the expression: $T_{r+1} = { }^5 C_r (2)^{5-r} (x^3)^{5-r} (-\frac{1}{3})^r (x^{-2})^r = { }^5 C_r (2)^{5-r} (-\frac{1}{3})^r x^{15-3r-2r} = { }^5 C_r (2)^{5-r} (-\frac{1}{3})^r x^{15-5r}$. To find the coefficient of $x^5$,we set the exponent of $x$ equal to $5$: $15 - 5r = 5 \implies 5r = 10 \implies r = 2$. Substituting $r = 2$ into the coefficient part: ${ }^5 C_2 (2)^{5-2} (-\frac{1}{3})^2 = 10 	imes 2^3 	imes \frac{1}{9} = 10 	imes 8 	imes \frac{1}{9} = \frac{80}{9}$.

The coefficient of $x^5$ in the expansion of $\left(2 x^3-\frac{1}{3 x^2}\right)^5$ is

Similar Questions

In the binomial expansion of $(a-b)^n, n \geq 5$,the sum of the $5^{\text{th}}$ and $6^{\text{th}}$ terms is zero. Then the value of $\frac{a}{b}$ is

Let $n$ be a positive even integer. If the ratio of the largest coefficient and the $2^{nd}$ largest coefficient in the expansion of $(1+x)^{n}$ is $11:10$, then the number of terms in the expansion of $(1+x)^{n}$ is:

For each positive integer $n$,let $A_n = \max \left\{ \binom{n}{r} \mid 0 \leq r \leq n \right\}$. Then,the number of elements $n \in \{1, 2, \ldots, 20\}$ for which $1.9 \leq \frac{A_n}{A_{n-1}} \leq 2$ is

If the second,third and fourth terms in the expansion of $(x+y)^{n}$ are $135$,$30$ and $\frac{10}{3}$,respectively,then $6(n^3+x^2+y)$ is equal to .............

Find the coefficient of $x^{5}$ in the expansion of $(x+3)^{8}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module